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Murljashka [212]
3 years ago
7

An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of

Chemistry
1 answer:
kherson [118]3 years ago
6 0

Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

MAss glucose = moles glucose* molecular weight of glucose

MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

We need 8.11 grams of glucose for this solution

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For each molecule, specify the polarity of the bonds and the overall polarity of the molecule.
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Describe in your own words, in terms of particle movement and energy, what occurs when a liquid is heated to its boiling point.
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3 years ago
A compound of mercury and oxygen is heated in order to decompose the compound. A 4.08 grams sample of mercury oxide upon heating
arlik [135]

Answer:

HgO (empirical formula)

Explanation:

4.08 - 3.78 = 0.3g (oxygen)

(\frac{4.08}{201})   \:  \:  \:  (\frac{0.3}{16} )

0.02 : 0.02

0.02/0.02 : 0.02/0.02

1 : 1 (ratio)

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7 0
2 years ago
How much heat is given off when 869 g iron bar cools from 94 degrees celsius to 5
melomori [17]

Answer:

Heat given off was -34.34kJ

Explanation:

Mass of iron bar = 869g

Initial temperature (T1) = 94°C

Final temperature (T2) = 5°C

Specific heat capacity of iron (c) = 0.444J/g°C

Heat energy (Q) = Mc∇T

Q = heat energy

c = specific heat capacity

∇T = change in temperature

M = mass of the substance

Q = mc∇T

∇T = T2 - T1

Q = Mc(T2 -T1)

Q = 869 * 0.444 * (5 - 94)

Q = 385.836 * -89

Q = -34339.404J

Q = -34.34kJ

The heat given of was -34.34kJ

4 0
3 years ago
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